Tricky integral

I just can't solve this:
[tex]I(\gamma)=\int_{-\infty}^{\infty} e^{-\sqrt{x^2+\gamma^2}}dx[/tex]
I have tried a trigonometric substitution
[tex]x=\gamma \tan(\theta)[/tex]
but I am not happy with the result :grumpy:.
Anyone got a hint?

Just go a bit further. You can get dx in terms of dy, y, and [tex]\gamma[/tex] by using the substitution and some algebra. By the way, the suggestion was to simplify the expontential. Hopefully, the resulting integrand will be more reasonable to deal with, such as integration by parts. I haven't solved it, but the substitution looks like it takes you in a promising direction.

I just can't solve this:
[tex]I(\gamma)=\int_{-\infty}^{\infty} e^{-\sqrt{x^2+\gamma^2}}dx[/tex]
I have tried a trigonometric substitution
[tex]x=\gamma \tan(\theta)[/tex]
but I am not happy with the result :grumpy:.
Anyone got a hint?

hotvette: I am not sure about this but I'll try. The pain is you have to split the integral because of the term [tex]\sqrt{y^2-\gamma^2}[/tex]

Pseudo Statistic: I do not want a numerical result this time... sorry.

Tom Mattson: No this is not a high school problem, any recommendation on where to post it? Good try with the log but integrals unfortunately do not have this property, I can not even think of any non trivial integral (that is the integral of 1 or that of 0) that would have this property.

The idea is to integrate the new integral with respect to [itex]x[/itex], then integrate that with respect to [itex]a[/itex], and then recognize that [itex]I(\gamma)=F(\gamma,1)[/itex].

Of course, I couldn't get it to work because of that [itex]x[/itex] in my [itex]du[/itex]. You can solve for [itex]x[/itex] in terms of [itex]u[/itex], but it gets ugly again. But maybe you can find a parameterization that works.

So far, I've figured out that since the integrand is an even function, instead of integrating from negative infinity to positive infinity you can integrate from 0 to positive infinity and multiply the result by two. Also, when gamma = 0, it's pretty easy to show that I(0) = 2*1 = 2. I haven't figured out the rest, though.

Thanks JoAuSc, it's a very wise thing to search for simplifications and particular cases. The "rest" is the tough part, though.

Anyway, I have the answer!
I just couldn't find it in text books but this integral is well known as the "modified Bessel function of second kind". As expected it does not have a primitive but that's all right because it is the answer to my original question.

Thanks all for helping out and sorry I had you sweat on such a tough problem.

A friend a mine is a researcher in Maths and eventhough he is not a specialist in integration, a friend of his just recognised the function right away! Like: "Hey dude, this looks like [tex]K_1(\gamma)[/tex], one of the modified Bessel function of second kind "... nevermind.

As this function is not usually defined by its integral (but by a more generic differential equation), you might not recognize it at first sight. The one we have actually corresponds to equation 7 in the second link, with n=1 and knowing that
[tex]\frac{1}{2}!=\frac{\sqrt{pi}}{2}[/tex]